select
ot.user_id,
count(distinct vt.visit_time) as visit_nums #对于同一用户多条下单记录匹配的多条访问结果去重
from order_tb ot
left join visit_tb vt on ot.user_id=vt.user_id
    and date(vt.visit_time)='2022-09-02'
where date(ot.order_time)='2022-09-02'
group by ot.user_id
order by visit_nums desc